Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
const int N = 3e3+3;
double dp[N][N], p[N];
int main() {
	int n;
	cin >> n;
	for(int i = 0; i < n; ++i){
		dp[i][0] = 1;
		cin >> p[i];
	}
	dp[n][0] = dp[n+1][0] = 1;
	for(int i = n; i > 0; --i)
		for(int heds = 1; heds <= n; ++heds){
			dp[i][heds] = p[i-1] * dp[i+1][heds-1];
			dp[i][heds] += (1 - p[i-1]) * dp[i+1][heds];
		}
	cout << setprecision(11) << dp[1][(n+2)/2];
	return 0;
}